How to show $\operatorname{Pic}(X)=0$? Exercise $14.2$.Q Vakil's notes

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I'm reading Vakil's notes and I'm struggling with the exercise $14.2$.Q. I've been able to prove everything except $\operatorname{Pic}(X)=0$ with $$ X=\operatorname{Spec}\frac{k[x,y,z]}{(xy-z^2)}. $$ I tried to find an affine covering where all the rings are UFDs, but without results. Also I tried to compute $\operatorname{H^1}(X,\mathcal{O}_X^{\times})$ using the Cech cohomology, but also here I don't know how to set my covering.

Some (maybe usefull) facts: this ring is normal and the class group of $X$ is $\mathbb{Z}/2\mathbb{Z}$.

Thank you in advance to those who can answer or suggest something.

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This is example 6.11.3 in Hartshorne. In 6.11.2 he argues that by the proof of Prop 6.11, Cartier divisors correspond to certain Weil divisors: if a Cartier divisor is given by $(U_i, f_i),$ then one associates a Weil divisor $\sum\limits_Y v_Y(f_i) [Y]$ to it. Weil divisors one gets this way are locally principal, meaning their restrictions on some open covering are principal.

Also note that by 6.15, $\operatorname{Pic}X \simeq \operatorname{CaCl}X.$

Now, the generator of $\operatorname{Cl}X \simeq \mathbb{Z}/2\mathbb{Z} $ is given by $V(z,y)$, which is not locally principal because $p:=(z,y)$ is not locally principal in the local ring of the point $(0,0,0) =: [m]$. Indeed, $m/m^2$ is a vector space spanned by $x, y$ and $z$, and $p \otimes \frac{k[x,y,z]}{(xy-z^2)}_m$ is its subspace which contains $y$ and $z$, so $p$ can't be principal in $\frac{k[x,y,z]}{(xy-z^2)}_m$.